energy conservation on charge sharing

/ (switch) ------ ------- | | | | = C1 = C2 | | | | Gnd Gnd C1, C2 are same capacitance C1 is precharged to Q1.(voltage is V1) when switch is closed, it will occur charge sharing. but i wonder the circuit apply to energy conservation law ( sorry I'm poor at english ;;) ,E = 1/2 * C * V^2 . when I calculated, it reduced half of original energy.

Asked By: baumpuppe
On: Aug 14, 2005 10:26:34 AM

Comments(5)

very true... When you see the voltage across the terminal then it will be equal to your supply voltage, whereas when you share the voltage shared by both the capacitors then the sum of voltage between the two capacitors is equal to the supply voltage when they are connected in parallel. http://www.sterling-energy.com/services/asset-acquisition-and-transition-support

Comment By: megna73
On: Oct 12, 2010 3:54:16 AM

E1=0.5*c*v*v . E2=0.5 *2*c * 0.5V*0.5v . e2 is infact 1/2 of e1.this amnt of energy is not conserved but spent to charge the other capacitor like a voltage supply spending energy to charge the cap.it is the number of charges that is conserved as charges cannot be created or destroyed.

Comment By: venkateshr
On: Aug 17, 2005 4:27:03 AM

sorry..this qna bbs can't be modify my article. fix (E = 0.5 * 2C * (0.5V)^2 * 2) to (E = 0.5 * C * (0.5V)^2 * 2 ) E means energy.(generally noted W)

Comment By: baumpuppe
On: Aug 15, 2005 7:14:17 PM

thank you for your answer. but I wanted to know that " E(energy) = 0.5 * C * V^2 " is correct for this case. you say Q is voltage. but I noted Q is "quantity of charge[Coulomb] " in this case, capacitor is connected to parallel. t= infinite , equivalent of this circuit will be double size of capacitor and half of voltage but same quantity of Q(charge). apply to energy formula,E=0.5*C*V^2 (relation to Q=CV) E = 0.5 * 2C * ( 0.5 V)^2 *2 -> it means half of voltage. = 0.5*CV^2 (at t= infinite) I dont know why this result is not same as E= CV^2 at t<0. plz answer to me.

Comment By: baumpuppe
On: Aug 15, 2005 6:44:12 PM

Lets say your Vdd=10V. so when the switch is open the capactior charges to 10V. but when the switch is closed charge sharing takes place. Since there is no loss in the circuit, the total charge will be: Q3(C1+C2)=Q1C1 + Q2C2 since both capacitors are equal: Let C1=C2=C 2C*Q3=C(Q1 + Q2) Q3 = Q1 + Q2/2 Now since charge of Q2 will rise exponentially the charge at t=0 will be 0V thus Q3 = Q1/2 =10/2 = 5V i hope this is what you meant

Comment By: rskurane
On: Aug 15, 2005 2:03:21 PM

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